TOWARDS the ALBERTSON CONJECTURE Dedicated to The

TOWARDS THE ALBERTSON CONJECTURE

JANOS´ BARAT´ AND GEZA´ TOTH´

Abstract. Albertson conjectured that if a graph G has chromatic number r then its crossing number is at least as much as the crossing number of Kr. Albertson, Cranston, and Fox veriﬁed the conjecture for r ≤ 12. In this note we prove it for r ≤ 16.

Dedicated to the memory of Michael O. Albertson.

1. Introduction Graphs in this paper are without loops and multiple edges. Every planar graph is four-colorable by the Four Color Theorem [2, 23]. Efforts to solve the Four Color Problem had a great effect on the development of graph theory, and it is one of the most important theorems of the field. The crossing number cr(G) of a graph G is the minimum number of edge crossings in a drawing of G in the plane. It is a natural relaxation of planarity, see [24] for a survey. The chromatic number χ(G) of a graph G is the minimum number of colors in a proper coloring of G. The Four Color Theorem states if cr(G) = 0 then χ(G) 4. Oporowski and Zhao [18] proved that every graph with crossing number at most≤ two is 5-colorable. Albertson et al. [5] showed that if cr(G) 6, ≤4 then χ(G) 6. It was observed by Schaefer that if cr(G) = k then χ(G) = O(√k) and this bound≤ cannot be improved asymptotically [4]. It is well-known that graphs with chromatic number r do not necessarily contain Kr as a subgraph, they can have clique number 2 [26]. The Hajós conjecture proposed that graphs with chromatic number r contain a subdivision of Kr. This conjecture, whose origin is unclear but attributed to Hajós, turned out to be false for r 7. Moreover, it was shown by Erd˝os and Fajtlowicz [9] that almost all graphs≥ are counterexamples. Albertson conjectured the following.

Conjecture 1. If χ(G) = r, then cr(G) cr(Kr). ≥ This statement is weaker than Haj´os’ conjecture, since if G contains a subdivision of Kr then cr(G) cr(Kr). For r = 5, Albertson’s≥ conjecture is equivalent to the Four Color Theorem. Oporowski and Zhao [18] veriﬁed it for r = 6, Albertson, Cranston, and Fox [4] proved it for r 12. In this note we take one more little step. ≤ Theorem 2. For r 16, if χ(G) = r, then cr(G) cr(Kr). ≤ ≥

Date: September 7, 2009. 2000 Mathematics Subject Classiﬁcation. Primary 05C10; Secondary 05C15. Research is supported by OTKA Grant PD 75837. Research is supported by OTKA T 038397 and T 046246. 1 2 J. BARAT´ AND G. TOTH´

In their proof, Albertson, Cranston, and Fox combined lower bounds for the number of edges of r-critical graphs, and lower bounds on the crossing number of graphs with given number of vertices and edges. Our proof is very similar, but we use better lower bounds in both cases. Albertson, Cranston, and Fox proved that any minimal counterexample to Al- bertson’s conjecture should have less than 4r vertices. We slightly improve this result as follows. Lemma 3. If G is an r-critical graph with n 3.57r vertices, then cr(G) ≥ ≥ cr(Kr). In Section 2 we review lower bounds for the number of edges of r-critical graphs, in Section 3 we discuss lower bounds on the crossing number, and in Section 4 we combine these bounds to obtain the proof of Theorem 2. In Section 5 we prove Lemma 3. The letter n always denotes the number of vertices of G. In notation and termi- nology we follow Bondy and Murty [6]. In particular, the join of two disjoint graphs G and H arises by adding all edges between vertices of G and H. It is denoted by G H. A vertex v is called simplicial if it has degree n 1. If a graph G contains a subdivision∨ of H, then we also say that G contains a topological− H. A vertex v is adjacent to a vertex set X means that each vertex of X is adjacent to v.

2. Color-critical graphs Around 1950, Dirac introduced the concept of color criticality in order to simplify graph coloring theory, and it has since led to many beautiful theorems. A graph G is r-critical if χ(G) = r but all proper subgraphs of G have chromatic number less than r. In what follows, let G denote an r-critical graph with n vertices and m edges. Since G is r-critical, every vertex has degree at least r 1 and therefore, 2m (r 1)n. Dirac [7] proved that for r 3, if G− is not complete, then 2m ≥(r −1)n + (r 3). For r 4, Dirac [8] gave≥ a characterization of r-critical ≥ − − ≥ graphs with excess r 3. For any fixed r 3 let ∆r be the family of graphs G − ≥ whose vertex set consists of three non-empty, pairwise disjoint sets A, B1,B2 with B1 + B2 = A + 1 = r 1 and two additional vertices a and b such that A and B| | B| both| span| | cliques− in G, they are not connected by any edge, a is connected 1 ∪ 2 to A B1 and b is connected to A B2. See Figure 1. Graphs in ∆r are called Hajós graphs∪ of order 2r 1. Observe∪ that that these graphs have chromatic number r − and they contain a topological Kr, hence they satisfy Hajós’ conjecture. Gallai [10] proved that r-critical graphs with at most 2r 2 vertices are the join of two smaller graphs, i.e. their complement is disconnected.− Based on this observation, he proved that non-complete r-critical graphs on at most 2r 2 vertices have much larger excess than in Dirac’s result. − Lemma 4. [10] Let r, p be integers satisfying r 4 and 2 p r 1. If G is an r-critical graph with n = r + p vertices, then 2m≥ (r 1)≤n + ≤p(r −p) 2, where ≥ − − − equality holds if and only if G is the join of Kr−p− and G ∆p . 1 ∈ +1 Since every G ∆p contains a topological Kp , the join of Kr−p− and G ∈ +1 +1 1 contains a topological Kr. This yields a slight improvement for our purposes. 3 a

Figure 1. The family ∆r

Corollary 5. Let r, p be integers satisfying r 4 and 2 p r 1. If G is an ≥ ≤ ≤ − r-critical graph with n = r + p vertices, and G does not contain a topological Kr, then 2m (r 1)n + p(r p) 1. ≥ − − − We call the bound given by Corollary 5 the Gallai bound. For r 3, let r denote the family of graphs G, whose vertex set consists of four ≥ E non-empty pairwise disjoint sets A1, A2,B1,B2, where B1 + B2 = A1 + A2 = r 1 and A + B r 1, and one additional vertex| c|such| that| |A =| A| |A − | 2| | 2| ≤ − 1 ∪ 2 and B = B1 B2 are cliques in G, NG(c) = A1 B1 and a vertex a A is adjacent to a vertex b∪ B if and only if a A and b ∪B . ∈ ∈ ∈ 2 ∈ 2 Clearly r ∆r, and every graph G r is r-critical with 2r 1 vertices. KostochkaE and⊃ Stiebitz [15] improved the bound∈ E of Dirac as follows. −

Lemma 6. [15] Let r 4 and G be an r-critical graph. If G is neither Kr nor a ≥ member of r, then 2m (r 1)n + (2r 6). E ≥ − − It is not diﬃcult to prove that any member of r contains a topological Kr. Indeed, A and B both span a complete graph on rE 1 vertices. We only have − to show that vertex c is connected to A2 or B2 by vertex-disjoint paths. To see this, we observe that A2 or B2 is the smallest of A1 , A2 , B1 , B2 . Indeed, if B was the smallest,| | then| A| > B and B {|> B| | implies| | | | A |}+ B > | 1| | 2| | 1| | 2| | 1| | 2| | 2| B1 + B2 = r 1 contradicting our assumption. We may assume that A2 is the smallest.| | | | Now −c is adjacent to A , and there is a matching of size A | between| 1 | 2| B1 and B2 and between B2 and A2, respectively. That is, we can ﬁnd a set S of disjoint paths from c to A . In this way A c S is a topological r-clique. 2 ∪ ∪ Corollary 7. Let r 4 and G be an r-critical graph. If G does not contain a ≥ topological Kr then 2m (r 1)n + (2r 6). ≥ − − 4 J. BARAT´ AND G. TOTH´ c

A1 B1

A2 B2

Figure 2. The family r E

Let us call this the Kostochka, Stiebitz bound, or KS-bound for short. In what follows, we obtain a complete characterization of r-critical graphs on r +3 or r + 4 vertices. Lemma 8. For r 8, there are precisely two r-critical graphs on r + 3 vertices. They can be constructed≥ from two 4-critical graphs on seven vertices by adding simplicial vertices.

Figure 3. The two 4-critical graphs on seven vertices

Proof. The proof is by induction on r. For the base case r = 8, there are precisely two 8-critical graphs on 11 vertices, see Royle’s complete search [21]. Let G be an r-critical graph with r 9 and n = r + 3 12. We know that the minimum degree is at least r 1 = n ≥4. If G has a simplicial≥ vertex v, then we use induction. So we may assume− that− every vertex in G, the complement of G has degree 1, 2 or 3. By Gallai’s theorem, G is disconnected. Observe the following: if there are at least four independent edges in G, then χ(G) n 4 = r 1, a ≤ − − contradiction. That is, there are at most three independent edges in G. Therefore, G has two or three components. If there is a triangle in the complement, then we can save two colors. If there were two triangles, then χ(G) n 4 = r 1, a contradiction. ≤ − − Assume that there are three components in G. Since each degree is at least one, there are at least three independent edges. Therefore, there is no triangle in G and 5 no path with three edges. That is, the complement consists of three stars. Since the degree is at most three and there are at least 12 vertices, there is only one possibility: G = K , K , K , , see Figure 4. 1 3 ∪ 1 3 ∪ 1 3

Figure 4. The complement and a removable edge

We have to check whether this concrete graph is indeed critical. We observe, that the edge connecting two centers of these stars is not critical, a contradiction. In the remaining case, G has two components H1 and H2. Since there are at most three independent edges, there is one in H1 and two in H2. It implies that H1 has at most four vertices. Therefore, H2 has at least eight vertices. Consider a spanning tree T of H2 and remove two adjacent vertices of T , one of them being a leaf. It is easy to see that the remainder of T contains a path with three edges. Therefore, in total we found three independent edges of H2, a contradiction. We need the following result of Gallai. Theorem 9. [10] Let r 3 and n < 5 r. Then every r-critical, n-vertex graph ≥ 3 contains at least 3 5 r n simplicial vertices. 2 3 − Lemma 10. For r 6, there are precisely twenty-two r-critical graphs on r+4 vertices. They can be constructed≥ by adding simplicial vertices to one of the following: a 3-critical graph on seven vertices, four 4-critical graphs on eight vertices, sixteen 5-critical graphs on nine vertices, or a 6-critical graphs on ten vertices. Proof. For the base of induction, we use Royle’s table again, see [21]. The full computer search shows that there are precisely twenty-two 6-critical graphs on ten vertices. For the induction step, we use Lemma 9 and see that there are at least r 6 simplicial vertices. Since r 7, there is always a simplicial vertex. We remove it− and use the induction hypothesis≥ to ﬁnish the proof. There is an explicit list of twenty-one 5-critical graphs on nine vertices [21]. We have checked, partly manually, partly using Mader’s extremal result [16], that each of those graphs contains a topological K5. Also the above mentioned 6-critical graph on ten vertices contains a topological K6. These results imply the following Corollary 11. Any r-critical graph on at most r + 4 vertices satisfy the Haj´os conjecture. 6 J. BARAT´ AND G. TOTH´

We conjecture that the following slightly more general statement can be proved with similar methods. Conjecture 12. Let G be an r-critical graph on r +o(r) vertices. Then G satisﬁes the Haj´os conjecture.

3. The crossing number It follows from Euler’s formula that a planar graph can have at most 3n 6 edges. Suppose that G has m 3n 6 edges. By deleting crossing edges one− by one, it follows by induction that≥ for n− 3, ≥ (1) cr(G) m 3(n 2) ≥ − − Pach et. al. [19] generalized it and proved the following lower bounds. Each one holds for any graph G with n 3 vertices and m edges. ≥ (2) cr(G) 7m/3 25(n 2)/3 ≥ − −

(3) cr(G) 3m 35(n 2)/3 ≥ − −

(4) cr(G) 4m 103(n 2)/6 ≥ − −

(5) cr(G) 5m 25(n 2) ≥ − − Inequality (1) is the best for m 4(n 1), (2) is the best for 4(n 2) m 5(n 2), (3) is the best for 5(n≤ 2) − m 5.5(n 2), (4) is the best− for≤ 5.5(≤n 2)− m 47(n 2)/6, and (5)− is the≤ best≤ for 47(n− 2)/6 m. It was− also≤ shown≤ in [19]− that (1) can not be improved in− the range≤ m 4(n 1), and (2) can not be improved in the range 4(n 2) m 5(n 2), apart≤ − from an additive constant. The other inequalities are− conjectured≤ ≤ to be− far from opti- mal. Using the methods in [19] one can obtain an infinite family of such linear inequalities, of the form am b(n 2). The most important inequality− for− crossing numbers is undoubtedly the Crossing Lemma, first proved by Ajtai, Chvátal, Newborn, Szemerédi[1], and independently by Leighton [13]. If G has n vertices and m 4n edges, then ≥ 1 m3 (6) cr(G) . ≥ 64 n2 1 The original constant was much larger, the constant 64 comes from the well-known probabilistic proof of Chazelle, Sharir, and Welzl [3]. The basic idea is to take a random spanned subgraph and apply inequality (1) for that. The order of magnitude of this bound can not be improved, see [19], the best known constant is obtained in [19]. If G has n vertices and m 103 n edges, then ≥ 16 1 m3 (7) cr(G) . ≥ 31.1 n2 The proof is very similar to the proof of (6), the main difference is that instead of (1), inequality (4) is applied for the random subgraph. The proof of the following technical lemma is based on the same idea. 7

Lemma 13. Suppose that n 10, and 0

cr(G) cr(n,m,p). ≥ Proof. Observe that inequality (4) does not hold for graphs with at most two vertices. For any graph G, let

cr(G) if n 3 ≥ ′  4 if n = 2 cr (G) =   18 if n = 1 35 if n = 0   It is easy to see that for any graph G

′ 103 (8) cr (G) 4m (n 2). ≥ − 6 − Let G be a graph with n vertices and m edges. Consider a drawing of G with cr(G) crossings. Choose each vertex of G independently with probability p, and let G′ be a subgraph of G spanned by the selected vertices. Consider the drawing of G′ inherited from the drawing of G, that is, each edge of G′ is drawn exactly as it is drawn in G. Let n′ and m′ be the number of vertices and edges of G′, and let x be the number of crossings in the present drawing of G′. Using that E(n′) = pn, E(m′) = p2m, E(x) = p4cr(G), and the linearity of expectations,

′ ′ ′ ′ ′ ′ E(x) E(cr(G )) E(cr (G )) 4P (n = 2) 18P (n = 1) 35P (n = 0) ≥ ≥ − − − ≥

103 103 n − − 4p2m pn + 4 p2(1 p)n 2 18np(1 p)n 1 35(1 p)n ≥ − 6 3 − 2 − − − − − ≥

103 103 − 4p2m pn + 5n2(1 p)n 2. ≥ − 6 3 − − Dividing by p4 we obtain the statement of the Lemma.

Note that in our applications p will be at least 1/2, n will be at least 13, therefore, 2 n−2 5n (1−p) the last term in the inequality, p4 , will be negligible. We also need some bounds on the crossing number of the complete graph, cr(Kr). It is not hard to see that

1 r r 1 r 2 r 3 (9) cr(Kr) Z(r) = − − − , ≤ 4 j2k 2 2 2 see e. g. [22]. Guy conjectured [11] that cr(Kr) = Z(r). This conjecture has been veriﬁed for r 12 but still open for r > 12. The best known lower bound is due to ≤ de Klerk et. al. [14]: cr(Kr) 0.86Z(r). ≥ 8 J. BARAT´ AND G. TOTH´

4. Proof of Theorem 2

Suppose that G is an r-critical graph. If G contains a topological Kr, then clearly cr(G) cr(Kr). Suppose in the sequel that G does not contain a topological Kr. Therefore,≥ we can apply the Kostochka, Stiebitz, and the Gallai bounds on the number of edges. Then we use Lemma 13 to get the desired lower bound on the crossing number. Albertson et. al. [4] used the same approach, but they used a weaker version of the Kostochka, Stiebitz, and the Gallai bounds, and instead of Lemma 13 they applied the weaker inequality (4). In the next table, we include the results of our calculations. For comparison, we also included the result Albertson et al. might have had using (4). In the Appendix we present our simple Maple program performing all calculations.

1. Let r = 13. By (9) we have cr(K ) 225. 13 ≤ n e bound (4) p ⌈cr(n, m, p)⌉ 18 128 238 0.719 288 19 135 249 0.732 296 20 141 255 0.751 298 21 146 258 0.774 294 If n 22, then the KS-bound combined with (4) gives the desired result. 2m 12≥n + 20 cr(G) 4(6n + 10) 103/6(n 2) 224.67, if n 22. ≥ ⇒ ≥ − − ≥ ≥ 2. Let r = 14. By (9) we have cr(K ) 315. 14 ≤ n e bound (4) p ⌈cr(n, m, p)⌉ 19 146 293 0.659 388 20 154 307 0.670 402 21 161 318 0.684 407 22 167 325 0.702 406 23 172 328 0.723 398 24 176 327 0.747 384 25 179 322 0.775 366 26 181 312 0.807 344 If n 27, then the KS-bound combined with (4) gives the desired result. 2m 13≥n + 22 cr(G) 4(6.5n + 11) 103/6(n 2) 316, if n 27. ≥ ⇒ ≥ − − ≥ ≥ 3. Let r = 15. By (9) we have cr(K ) 441. 15 ≤ n e bound (4) p ⌈cr(n, m, p)⌉ 20 165 351 0.610 510 21 174 370 0.617 531 22 182 385 0.623 542 23 189 396 0.642 545 24 195 403 0.659 539 25 200 406 0.678 526 26 204 404 0.700 508 27 207 399 0.725 484 Suppose now that G is 15-critical and n 28. By the KS-bound we have m 7n + 12. Apply Lemma 13 with p = 0.764≥ and a straightforward calculation gives≥ cr(G) cr(n, m, 0.764) 441. ≥ ≥ 9

4. Let r = 16. By (9) we have cr(K ) 588. 16 ≤ n e bound (5) p ⌈cr(n, m, p)⌉ 21 185 450 0.567 657 22 195 475 0.573 687 23 204 495 0.581 706 24 212 510 0.592 714 25 219 520 0.605 712 26 225 525 0.621 701 27 230 525 0.639 683 28 234 520 0.659 658 29 237 510 0.681 628 30 239 495 0.706 593 31 246 505 0.713 601 Suppose now that G is 16-critical and n 32. By the KS-bound we have m 7.5n + 13. Apply Lemma 13 with p =≥ 0.72 and again a straightforward calculation≥ gives cr(G) cr(n, m, 0.72) 588. This concludes the proof≥ of Theorem 2.≥

Remark. For r 17 we could not completely verify Albertson’s conjecture. The next table contains≥ our calculations for r = 17. There are three cases, n = 32, 33, 34, for which our approach is not suﬃcient. By (9) we have cr(K ) 784. 17 ≤ n e bound from p bound using equation 5 cr(n, e, p) 22 206 530 0.530 832 23 217 560 0.534 874 24 227 585 0.541 902 25 236 605 0.550 917 26 244 620 0.560 920 27 251 630 0.573 913 28 257 635 0.588 897 29 262 635 0.604 872 30 266 630 0.622 840 31 269 620 0.643 802 32 271 605 0.665 759 33 278 615 0.672 765 34 286 630 0.677 779

Lemma 14. Let G be a 17-critical graph on n vertices. If n 35, then cr(G) 784 cr(K ). ≥ ≥ ≥ 17 Proof. Let p = 0.681. Then cr(G) cr(n, m, 0.681) 14.64n+280.38. Therefore, 784−280.38 ≥ ≥ if n 14.64 34.4, then we are done. (Without the probabilistic argument, the same≥ result holds≥ with n 44.) ≥ Lemma 15. Let G be a 17-critical graph on 32 vertices. Then cr(G) cr(K ). ≥ 17 Proof. Gallai [10] proved that any r-critical graph on at most 2r 2 vertices is a join of two smaller critical graphs. This is a structural version of the− Gallai bound. In our case, r = 17, and n = 2r 2 = 32. Assume that G = G G , where G is − 1 ∨ 2 1 r1-critical on n1 vertices, G2 is r2-critical on n2 vertices, where 17 = r1 + r2 and 10 J. BARAT´ AND G. TOTH´

32 = n1 + n2. The sum of the degrees of G can be estimated as the sum of the degrees of the vertices in Gi, for i = 1, 2, plus twice the number of edges between G1 and G2: 2m (r1 1)n1 + (r2 1)n2 + 2(r 3) + 2n1n2. How much do≥ we gain− with this− calculation compared− to the direct application of the Gallai bound on G? That is seen after a simple subtraction: (r 1)n +(r 1)n +2(r 3)+2n n (r 1)n 2(r 3) = (n r )n +(n r )n . 1− 1 2− 2 − 1 2− − − − 1− 1 2 2− 2 1 This value is minimal if n2 = r2 = 1. In that case, we gain n1 r1 = 15. That is, in our calculation we can add 15/2 edges, after which cr(G)− 834 arises. ⌈ ⌉ ≥ It is clear that our improvement on Gallai’s result relies on the fact that Kos- tochka and Stiebitz improved Dirac’s result.

5. Proof of Lemma 3 Suppose that r 17 and G is an r-critical graph with n vertices and m edges. If n 4r then the≥ statement holds by [4]. Suppose that 3.57r n 4r. In order to estimate≥ the crossing number of G, instead of the probabilistic≤ argument≤ in the proof of Lemma 13, we apply inequality (4) for each spanned subgraph of G with n exactly 52 vertices. Let k = 52 and let G1, G2,...,Gk be the spanned subgraphs of G with 52 vertices. Suppose that Gi has mi edges. Then for any i, by (4) we have 103 cr(Gi) 4mi 50, ≥ − 6 · consequently, k 1 103 4m n 2 50 103 n cr(G) 4mi 50 = − = ≥ n−4 − 6 · n−4 50 − n−4 6 52 48 Xi=1 48 48 4(n 2)(n 3)m 103 n(n 1)(n 2)(n 3) = − − − − − = 50 49 − 6 52 51 49 2(n 2)(n · 3)n(r 1) 103 n(n ·1)(n· 2)(n 3) − − − − − − = ≥ 50 49 − 6 52 51 49 n(n· 2)(n 3) r 1 103(n· 1)· = − − − − 49 25 − 6 52 51 n−· 4 · since we counted each possible crossing at most 48 times, and each edge of G n−2 exactly 50 times. Finally, some calculation shows that it is greater than 1 r(r 1)(r 2)(r 3) > cr(Kr) 64 − − − which proves the lemma. 2 Remarks 1. As we have already mentioned, see (7), the best known constant in the Crossing Lemma 1/31.1 is obtained in [19]. Montaron [17] managed to improve it slightly for dense graphs, that is, in the case when m = O(n2). His calculations are similar to the proof of Lemmas 3 and 13. 2. Our attack of the Albertson conjecture is based on the following philosophy. We calculate a lower bound for the number of edges of an r-critical n-vertex graph G. Then we substitute this into the lower bound given by Lemma 13. Finally, we compare the result and the Zarankiewicz number Z(r). For large r, this method 11 is not suﬃcient, but it gives the right order of magnitude, and the constants are roughly within a factor of 4. Let G be an r-critical graph with n vertices, where r n 3.57r. Then 2m (r 1)n. We can apply (7): ≤ ≤ ≥ − 1 ((r 1)n/2)3 (r 1)3n 1 Z(r) cr(G) − = − r(r 1)3 . ≥ 31.1 n2 31.1 8 ≥ 250 − ≥ 4 · 3. Let G = G(n, p) be a random graph with n vertices and edge probability p = p(n). It is known (see [12]) that there is a constant C0 > 0 such that if np > C0 then asymptotically almost surely we have np χ(G) < . log np Therefore, asymptotically almost surely n4p4 cr(Kχ(G)) Z(χ(G)) < . ≤ 64 log4 np On the other hand, by [20], if np > 20 then almost surely n4p2 cr(G) . ≥ 20000

Consequently, almost surely we have cr(G) > cr(Kχ(G)), that is, roughly speaking, unlike in the case of the Hajós conjecture, a random graph almost surely satisfies the statement of the Albertson conjecture. 4. If we do not believe in Albertson’s conjecture, we have to look for a counterexample in the range n 3.57r. Any candidate must also be a counterexample for the Hajós Conjecture. It≤ is tempting to look at Catlin’s graphs. k Let C5 denote the graph arising from C5 by repeating each vertex k times. That is, each vertex of C5 is blown up to a complete graph on k vertices and any edge of C5 is blown up to a complete bipartite graph Kk,k. Lemma 16. Catlin’s graphs satisfy the Albertson conjecture. Proof. It is known that χ(Ck) = 5 k . To draw Ck, there must be two copies of 5 ⌈ 2 ⌉ 5 K2k, a Kk and three copies of Kk,k drawn. Therefore

4 4 k 1 4 1 k k 4 cr(C ) 2Z(2k) + Z(k) + 3cr(Kk,k) 2 k + + 3 > 0.70k . 5 ≥ ∼ 4 4 2 2 On the other hand

4 1 5 4 (10) cr(K k ) cr(K 5 ) k < 0.62k χ(C5 ) ∼ 2 k ≤ 4 4 which proves the claim.

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Appendix start:=proc(r,n) local p,m,eredm,f,g,h,cr; if (n<=2*r-2) then p:=n-r; m:=ceil(((r-1)*n+p*(r-p)-1)/2); else m:=ceil(((r-1)*n+2*(r-3))/2); fi; 13 g:= ceil(5*m-25*(n-2)); print(m,g); f:= 4*m*x^2-(103/6)*n*x^3+(103/3)*x^4; eredm:=[solve((diff(f,x)/x)=0, x)]; print(evalf(eredm)); cr := min(eredm[1], eredm[2]); print(evalf(1/cr)); h:= f-(5*n^2*(1-1/x)^(n-2))/(1/x)^4; evalf((subs(x=cr, h))); end:

Department of Computer Science and Systems Technology, University of Pannonia, Egyetem u. 10, 8200 Veszprem,´ Hungary

Renyi´ Institute, Realtanoda´ u. 13-15, 1052 Budapest, Hungary